Prove $\int_0^1 \frac{4\cos^{-1}x}{\sqrt{2x-x^2}}\,dx=\frac{8}{9\sqrt{\pi}}\left(9\Gamma(3/4)^2{}_4F_3(\cdots)+\Gamma(5/4)^2{}_4F_3(\cdots)\right)$
About the formula in the headline given by Mathematica.
For the proof of the formula in the headline we have to know that for $\,\displaystyle |x|\leq \frac{1}{16}\,$ exists :
$\displaystyle f(x):=\sum\limits_{k=0}^\infty \frac{x^k}{4k+1}{\binom {4k} {2k}} / {\binom {k+\frac{1}{4}} {\frac{1}{2}} }={ \binom {\frac{1}{4}} {\frac{1}{2}}}^{-1} {}_4F_3\left( \begin{array}{c}\tfrac14\,\tfrac14\,\tfrac34\,\tfrac34\\\tfrac12\,\tfrac54\,\tfrac54\end{array};16x\right)$
$\displaystyle g(x):=\sum\limits_{k=0}^\infty \frac{x^k}{4k+3}{\binom {4k+2} {2k+1}} / {\binom {k+\frac{3}{4}} {\frac{1}{2}} }=\frac{2}{3} { \binom {\frac{3}{4}} {\frac{1}{2}}}^{-1} {}_4F_3\left( \begin{array}{c}\tfrac34\,\tfrac34\,\tfrac54\,\tfrac54\\\tfrac32\,\tfrac74\,\tfrac74\end{array};16x\right)$
On the other hand we have (for the original integral divided by $\,4\,$):
$\enspace\displaystyle \int\limits_0^1 \frac{\arccos x}{\sqrt{2x-x^2}}dx = \int\limits_0^{\pi/2} \frac{t\sin t }{\sqrt{2\cos t-(\cos t)^2}}dt =$
$\displaystyle =2\int\limits_0^{\pi/2} \arcsin(\sqrt{\frac{\cos t}{2}})dt = \sqrt{2}\sum\limits_{k=0}^\infty \binom {2k} k \frac{1}{8^k(2k+1)} \int\limits_0^{\pi/2} \sqrt{\cos t}^{2k+1}dt$
$\displaystyle =\sqrt{2}\sum\limits_{k=0}^\infty \frac{1}{8^k(2k+1)} {\binom {2k} k}/{\binom {\frac{k}{2}+\frac{1}{4}} {\frac{1}{2}}}=\sqrt{2}\Big(f\big(\frac{1}{8^2}\big)+\frac{1}{8} g\big(\frac{1}{8^2}\big)\Big) $
Since $\,\,\displaystyle \frac{1}{\sqrt{2}} {\binom {\frac{1}{4}} {\frac{1}{2}}}^{-1} = \frac{1}{\sqrt{\pi}} \Gamma\left(\frac{3}{4}\right)^2\,$ and $\,\,\displaystyle \frac{1}{2\sqrt{2}} {\binom {\frac{3}{4}} {\frac{1}{2}}}^{-1} = \frac{2}{3\sqrt{\pi}} \Gamma\left(\frac{5}{4}\right)^2\,$ ,
we get what Mathematica creates.
Notes:
The formula for $f$ (above) is based on
$\displaystyle \frac{1}{4k+1} \binom {4k} {2k} / \binom {k+\frac{1}{4}} {\frac{1}{2}} = {\binom {\frac{1}{4}} {\frac{1}{2}} }^{-1} \frac{ ( \frac{1}{4} )_k (\frac{1}{4})_k (\frac{3}{4})_k (\frac{3}{4})_k }{(\frac{1}{2})_k (\frac{5}{4})_k (\frac{5}{4})_k } \frac{16^k}{k!}$
e.g. proofed by induction :
$k=0$ : o.k.
$k\to k+1$ :
$\displaystyle\frac{ \frac{1}{4k+3} \binom {4k+4} {2k+2} / \binom {k+\frac{5}{4}} {\frac{1}{2}} }{\frac{1}{4k+1} \binom {4k} {2k} / \binom {k+\frac{1}{4}} {\frac{1}{2}}} = \frac{2 (4k+1)^2 (4k+3)^2}{(k+1) (2k+1) (4k+5)^2} = \frac{{\binom {\frac{1}{4}} {\frac{1}{2}} }^{-1} \frac{ ( \frac{1}{4} )_{k+1} (\frac{1}{4})_{k+1} (\frac{3}{4})_{k+1} (\frac{3}{4})_{k+1} }{(\frac{1}{2})_{k+1} (\frac{5}{4})_{k+1} (\frac{5}{4})_{k+1} } \frac{16^{k+1} }{(k+1)!}}{ {\binom {\frac{1}{4}} {\frac{1}{2}} }^{-1} \frac{ ( \frac{1}{4} )_k (\frac{1}{4})_k (\frac{3}{4})_k (\frac{3}{4})_k }{(\frac{1}{2})_k (\frac{5}{4})_k (\frac{5}{4})_k } \frac{16^k}{k!}}$
The formula for $g$ (above) is based on
$\displaystyle \frac{1}{4k+3} \binom {4k+2} {2k+1} / \binom {k+\frac{3}{4}} {\frac{1}{2}} = \frac{2}{3}{\binom {\frac{3}{4}} {\frac{1}{2}} }^{-1} \frac{ ( \frac{3}{4} )_k (\frac{3}{4})_k (\frac{5}{4})_k (\frac{5}{4})_k }{(\frac{3}{2})_k (\frac{7}{4})_k (\frac{7}{4})_k } \frac{16^k}{k!}$
e.g. proofed by induction :
$k=0$ : o.k.
$k\to k+1$ :
$\displaystyle\frac{ \frac{1}{4k+7} \binom {4k+6} {2k+3} / \binom {k+\frac{7}{4}} {\frac{1}{2}} }{\frac{1}{4k+3} \binom {4k+2} {2k+1} / \binom {k+\frac{3}{4}} {\frac{1}{2}}} = \frac{2 (4k+3)^2 (4k+5)^2}{(k+1) (2k+3) (4k+7)^2 } = \frac{\frac{2}{3} {\binom {\frac{3}{4}} {\frac{1}{2}} }^{-1} \frac{ ( \frac{3}{4} )_{k+1} (\frac{3}{4})_{k+1} (\frac{5}{4})_{k+1} (\frac{5}{4})_{k+1} }{(\frac{3}{2})_{k+1} (\frac{7}{4})_{k+1} (\frac{7}{4})_{k+1} } \frac{16^{k+1} }{(k+1)!}}{ \frac{2}{3} {\binom {\frac{3}{4}} {\frac{1}{2}} }^{-1} \frac{ ( \frac{3}{4} )_k (\frac{3}{4})_k (\frac{5}{4})_k (\frac{5}{4})_k }{(\frac{3}{2})_k (\frac{7}{4})_k (\frac{7}{4})_k } \frac{16^k}{k!}}$